If $X\sim\chi^2_{6}$, what is the probability density function of $T = \frac{(X-6)}{\sqrt12}$?
The problem I'm confronted with is that the chi-squared random variable, $X$, can assume only positive values while the random variable $T$ can assume both positive and negative values.
I tried to use both the PDF technique and the CDF technique for transformations of random variables but in the end the PDF of $T$ could assume only positive values, that is my problem.
$X \sim \chi^2_{k}$ is a random variable with mean $k$ and variance $2k$, and so $T$ is a simply a "unitized" version of $X$, meaning $E[T] = 0, \operatorname{var}(T) = 1$. (Parenthetical remark: I would have loved to refer to this process as "normalization" but the risk of being misunderstood is too great!). Now, $X$ is also a Gamma random variable with order parameter $\frac{k}{2}$ and scale parameter $\frac{1}{2}$, that is, $$f_X(x) = \frac{\frac{1}{2}\left(\frac{x}{2}\right)^{k/2-1}e^{-x/2}}{\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{x\in (0,\infty)}~ = ~ \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{x\in (0,\infty)}$$ and since the transformation $X\to T = (X-\mu)/\sigma$ is a linear function, we have that $$f_T(y) = \sigma f_X(\sigma y+\mu) =\sqrt{2k}\frac{\left(\sqrt{2k}y+k\right)^{k/2-1}e^{-(\sqrt{2k}y+k)/2}}{2^{k/2}\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{y\in (-\sqrt{k/2},\infty)}$$ which for the case $k=6$ simplifies to $$\begin{align} f_T(y) &=\sqrt{12}\frac{\left(\sqrt{12}y+6\right)^{2}e^{-(\sqrt{12}y+6)/2}}{2^3\Gamma\left(3\right)}\mathbf 1_{y\in (-\sqrt{3},\infty)}\\ &= \frac{\sqrt{3}\left(\sqrt{3}y+3\right)^{2}e^{-(\sqrt{3}y+3)}}{2}\mathbf 1_{y\in (-\sqrt{3},\infty)} \end{align}$$ which agrees with @COOLSerdash's machine-computed answer.
